Solid Ba(NO₃)₂ is gradually dissolved in a 1.0 × 10^{-3} ,M ,Na₂CO₃ solution. At what

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6-2.Equilibrium-II (Ionic Equilibrium)

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Solid $Ba(NO_3)_2$ is gradually dissolved in a $1.0 \times 10^{-3} \,M \,Na_2CO_3$ solution. At what concentration of $Ba^{2+}$ will be precipitate begin to form ($K_{sp}$ for $BaCO_3 = 5.1 \times 10^{-9}$)

A

$4.1 \times 10^{-5} \,M$

B

$5.1 \times 10^{-6} \,M$

C

$8.1 \times 10^{-8} \,M$

D

$8.1 \times 10^{-7} \,M$

Solution

$5.1 \times 10^{-9}=\left[\mathrm{Ba}^{+2}\right]\left[10^{-3}\right] \Rightarrow\left[\mathrm{Ba}^{+2}\right]$

$=5.1 \times 10^{-6} \,\mathrm{m}$

Standard 11

Chemistry

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